Question

In an equilateral triangle ABC, D is a point on side BC such that BD =
1/3BC. Prove that 9 AD2

= 7AB2​

Answer 1

$\large\underline\mathfrak\red{Correct \: Question \: :- }$

• In an equilateral triangle ABC,D is a point on side BC such that BD = 1/3 BC. Prove that 9AD² = 7AB²

∆ABC is an Equilateral triangle

such that,

AB = BC = AC

Draw an altitude AE

Such that,

E lies on BC and AE perpendicular to BC

In ∆AEB

By Pythagoras theorem,

AB² = AE² + EB² ...(1)

In ∆AED

By Pythagoras theorem,

AD² = AE² + ED² ...(2)

From (1) and (2)

AD² = ED² + AB² - EB² ...(3)

Since,

$\bf\implies{EB = \frac{1}{2}BC = \frac{AB}{2} = \frac{3AB}{6} }$

And

$\bf\implies{ED = \frac{BC}{6} = \frac{AB}{6} }$

$\bf\implies{AD {}^{2} = \frac{AB {}^{2} }{36} + AB {}^{2} - \frac{9AB {}^{2} }{36} }$

Hence,

9AD² = 7AB²

$\huge\bf\underline\orange{Hence \: Proved \: ! }$

Answer 2

Answer:

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